A closed organ pipe is filled with water by one-fifth of its original volume. The percentage change in its fundamental frequency is:
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Question

A closed organ pipe is filled with water by one-fifth of its original volume. The percentage change in its fundamental frequency is:

G1 Physics · Accepted Answer

Fundamental frequency of a closed organ pipe is f = v/(4L). If water fills one-fifth of the pipe volume, for uniform cross-section it occupies one-fifth of the length. The remaining air-column length is L' = L - L/5 = 4L/5. New fundamental frequency: f' = v/(4L') = v/[4(4L/5)] = 5v/(16L). Original frequency: f = v/(4L) = 4v/(16L). Thus, Δf = f'-f = v/(16L). Percentage increase: (Δf/f)×100 = [v/(16L)]/[v/(4L)]×100 = 1/4×100 = 25%. Correct option: C.